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Math Help - Slope of the tangent line

  1. #1
    Junior Member xterminal01's Avatar
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    Slope of the tangent line

    The slope of the tangent live to a curve is given by f`(x)=3x^2+5 if the point
    (1,2) is on the curve, find the equation of the curve.
    if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
    So how would i go about finding the constant ?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by xterminal01 View Post
    The slope of the tangent live to a curve is given by f`(x)=3x^2+5 if the point
    (1,2) is on the curve, find the equation of the curve.
    if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
    So how would i go about finding the constant ?
    plug in the initial condition ...

    f(x) = x^3 + 5x + C

    f(1) = 1^3 + 5(1) + C

    2 = 6 + C

    C = -4
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by xterminal01 View Post
    The slope of the tangent live to a curve is given by f`(x)=3x^2+5 if the point
    (1,2) is on the curve, find the equation of the curve.
    if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
    So how would i go about finding the constant ?
    (1,2) <---this is your initial condition


    (1,2)\text{ on f(x) }\Rightarrow2=(1)^3+5(1)+C
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